Answer:
Correct option: c) the t distribution with 49 degrees of freedom must be used
Explanation:
The (1 - α)% confidence interval for population mean (μ) is:
[tex]CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]
OR
[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\frac{s}{\sqrt{n}}[/tex]
A t-interval is used when the there is no information provided about the population standard deviation and when the population is normally distributed.
In this case there is no information about the population standard deviation and the population is also not normally distributed.
But as the t-distribution is derived from the normal distribution, to construct a t-interval the sample drawn must be large, i.e. n > 30.
Because for large sample sizes the sampling distribution of sample means will follow a Normal distribution with mean μ of the population and standard deviation [tex]\frac{s}{\sqrt{n}}[/tex].
In this case the sample size is, n = 50 > 30.
So a t distribution with n - 1 = 50 - 1 = 49 degrees of freedom will be used to construct the confidence interval for mean.
